Difference between revisions of "1973 USAMO Problems/Problem 5"
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where <math>m</math>, <math>n</math> are distinct integers, and <math>d</math> is the common difference in the progression. Then we have | where <math>m</math>, <math>n</math> are distinct integers, and <math>d</math> is the common difference in the progression. Then we have | ||
| − | + | \[ \begin{array}{rcl} nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}} | |
| − | + | n^{3}q-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}+3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-m^{3}r&=&(n-m)^{3}p | |
| − | + | 3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}&=&(n-m)^{3}p+m^{3}r-n^{3}q | |
| − | + | (3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})(mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q | |
| − | + | nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}} | |
| − | + | mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}&=&(m-n)p^{\dfrac{1}{3}} | |
| − | + | (3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q | |
| − | + | q^{\dfrac{1}{3}}r^{\dfrac{1}{3}}p^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)} | |
| − | + | (pqr)^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)} \end{array}\] | |
Because <math>p</math>, <math>q</math>, <math>r</math> are distinct primes, <math>pqr</math> is not a perfect cube. Thus, the LHS is irrational but the RHS is rational, which is a contradiction. So, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression. | Because <math>p</math>, <math>q</math>, <math>r</math> are distinct primes, <math>pqr</math> is not a perfect cube. Thus, the LHS is irrational but the RHS is rational, which is a contradiction. So, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression. | ||
Revision as of 20:21, 20 June 2013
Problem
Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression.
Solution
Assume that the cube roots of three distinct prime numbers can be three terms of an arithmetic progression. Let the three distinct prime numbers be 
, 
, and 
WLOG, let 
Then,
![\[q^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+md\]](http://latex.artofproblemsolving.com/4/a/0/4a071d3477ec5ceddeea667cf58e0f20682d28f9.png)
![\[r^{\dfrac{1}{3}}=p^{\dfrac{1}{3}}+nd\]](http://latex.artofproblemsolving.com/c/f/8/cf8594129e79769cafaee3b4da58837178f92b46.png)
where 
, 
are distinct integers, and 
is the common difference in the progression. Then we have
\[ \begin{array}{rcl} nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}}
n^{3}q-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}+3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-m^{3}r&=&(n-m)^{3}p
3nm^{2}q^{\dfrac{1}{3}}r^{\dfrac{2}{3}}-3n^{2}mq^{\dfrac{2}{3}}r^{\dfrac{1}{3}}&=&(n-m)^{3}p+m^{3}r-n^{3}q
(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})(mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q
nq^{\dfrac{1}{3}}-mr^{\dfrac{1}{3}}&=&(n-m)p^{\dfrac{1}{3}}
mr^{\dfrac{1}{3}}-nq^{\dfrac{1}{3}}&=&(m-n)p^{\dfrac{1}{3}}
(3nmq^{\dfrac{1}{3}}r^{\dfrac{1}{3}})((m-n)p^{\dfrac{1}{3}})&=&(n-m)^{3}p+m^{3}r-n^{3}q
q^{\dfrac{1}{3}}r^{\dfrac{1}{3}}p^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)}
(pqr)^{\dfrac{1}{3}}&=&\dfrac{(n-m)^{3}p+m^{3}r-n^{3}q}{(3mn)(m-n)} \end{array}\]
Because , , 
are distinct primes, 
is not a perfect cube. Thus, the LHS is irrational but the RHS is rational, which is a contradiction. So, the cube roots of three distinct prime numbers cannot be three terms of an arithmetic progression.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
| 1973 USAMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
